http://google.com, pub-7771400403364887, DIRECT, f08c47fec0942fa0 Find the image of the circle |z-i|=1 under the mapping of f(z)= 1/2

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Find the image of the circle |z-i|=1 under the mapping of f(z)= 1/2


Solution
 
f(z)=(x-y)/(x²+y²)=x/(x²-y²)-it/(x²+y²)

Therefore 

U=x/(x²+y²), V=-y/(x²+y²)

Equivalently 

x=u/(u²-v²), y=v/(u²+v²)

|z-i|=1 is 
1/(u²+v²)-2u/(u²+v²)=0

If |z-1|=1 then 

|x+iy-1|=|x-1+it|=1

Therefore (x-1)²+y²=1

x²-2x+y²=0

Therefore 
1/(u²+v²)²-2u/(u²+v²)=0

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